Find $\lim_{x\to 1^+}(x-1)^{^{{(x-1)}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $0$ (Choice C) C $-1$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=1$ into $(x-1)^{^{{(x-1)}}}$ results in the indeterminate form $0^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(x-1)^{^{{(x-1)}}}$, we will find $\lim_{x\to 1^+}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 1^+}y$. $\ln(y) =\dfrac{\ln(x-1)}{(x-1)^{-1}}$ Substituting $x=1$ into $\dfrac{\ln(x-1)}{(x-1)^{-1}}$ results in the indeterminate form $\dfrac{-\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 1^+}\ln(y) \\\\ &=\lim_{x\to 1^+}\dfrac{\ln(x-1)}{(x-1)^{-1}} \\\\ &=\lim_{x\to 1^+}\dfrac{\dfrac{d}{dx}[\ln(x-1)]}{\dfrac{d}{dx}\left[(x-1)^{-1}\right]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 1^+}\dfrac{\left(\dfrac{1}{x-1}\right)}{-(x-1)^{-2}} \\\\ &=\lim_{x\to 1^+}[-(x-1)] \gray{\text{Substitution}} \\\\ &=0 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 1^+}\dfrac{\dfrac{d}{dx}[\ln(x-1)]}{\dfrac{d}{dx}\left[(x-1)^{-1}\right]}$ actually exists. We found that $\lim_{x\to 1^+}\ln(y)=0$, which means $\lim_{x\to 1^+}y=1$. [Why?] In conclusion, $\lim_{x\to 1^+}(x-1)^{^{{(x-1)}}}=1$.